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3 years ago

A reflection about the y axis is determined by what?

Mathematics

1 answer:

murzikaleks [220]3 years ago

6 0

Answer:

It is determined by the graph and what it is.

Step-by-step explanation:

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If QR= 3x-2, RS= 2x+1, QS=6x-3, and R is between Q and S, then RS=

rewona [7]

Answer:

RS = 5

Step-by-step explanation:

Since R is between Q and S , then

QS = QR + RS , that is

6x - 3 = 3x - 2 + 2x + 1

6x - 3 = 5x - 1 ( subtract 5x from both sides )

x - 3 = - 1 ( add 3 to both sides )

x = 2

Then

RS = 2x + 1 = 2(2) + 1 = 4 + 1 = 5

4 0

3 years ago

Find the number that makes the ratio equivalent to 5:6. :66

choli [55]

Answer:

55 - you multiplied 6 by 11 to get 66 so you multiply by 5 by 11 to get 55.

4 0

2 years ago

Picture shows question

GalinKa [24]

All work is show in the picture below.

5 0

3 years ago

Use the Fundamental Theorem of Calculus to find the "area under curve" of

lozanna [386]

Answer:

$\displaystyle A = 300$

General Formulas and Concepts:

Calculus

Integrals

Definite Integrals
Area under the curve
Integration Constant C

Integration Rule [Reverse Power Rule]: $\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C$

Integration Rule [Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)$

Integration Property [Multiplied Constant]: $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

Integration Property [Addition/Subtraction]: $\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx$

Area of a Region Formula: $\displaystyle A = \int\limits^b_a {[f(x) - g(x)]} \, dx$

Step-by-step explanation:

Step 1: Define

Identify

f(x) = 6x + 19

Interval [12, 15]

Step 2: Find Area

Substitute in variables [Area of a Region Formula]: $\displaystyle A = \int\limits^{15}_{12} {(6x + 19)} \, dx$
[Integral] Rewrite [Integration Property - Addition/Subtraction]: $\displaystyle A = \int\limits^{15}_{12} {6x} \, dx + \int\limits^{15}_{12} {19} \, dx$
[Integrals] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle A = 6\int\limits^{15}_{12} {x} \, dx + 19\int\limits^{15}_{12} {} \, dx$
[Integrals] Integrate [Integration Rule - Reverse Power Rule]: $\displaystyle A = 6(\frac{x^2}{2}) \bigg| \limits^{15}_{12} + 19(x) \bigg| \limits^{15}_{12}$
Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: $\displaystyle A = 6(\frac{81}{2}) + 19(3)$
Simplify: $\displaystyle A = 300$